# An unusual equivalent form of Riemann hypothesis

Let $$G(x)=\sum_{k\leq x}\frac{\mu(k)}{k}$$, where $$\mu$$ is the Mobius function. From this question and its answer, its mention the Riemann hypothesis is equivalent to $$G(x)=O(x^{-\frac{1}{2}+\epsilon})$$.

I have never heard of this, the closet thing I have heard is the Riemann hypothesis is equivalent to $$M(x)=O(x^{\frac{1}{2}+\epsilon})$$ for all $$\epsilon>0$$, where $$M(x)=\sum_{k\leq x}\mu(k)$$, the Mertens function. I have tried to prove the above fact but did not succeed. So I would like to ask for reference for the above equivalent form of the Riemann hypothesis(the $$G(x)$$ one), better to contain a proof of it.

• Abel's summation formula gives $$\sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} + \int_1^x {\frac{{M(t)}}{{t^2 }}{\rm d}t} .$$ Can you finish from here?
– Gary
Commented Mar 16, 2023 at 11:29
• Can you write an answer? I try to bounded the integral term, but then the integral seems unbounded when $\epsilon>0$, also there are constant term that didn't cancel out. Commented Mar 16, 2023 at 12:06
• Some explicit bounds can be found at the link below. One such bound is $$|M(t)|<\frac{t}{\log(t)^{11/9}}$$ for $x>1000$. en.wikipedia.org/wiki/Mertens_function#Known_upper_bounds Commented Mar 16, 2023 at 12:17

Abel's summation formula gives $$G(x):=\sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} + \int_1^x {\frac{{M(t)}}{{t^2 }}{\rm d}t} .$$ Assume RH. Then $$M(x)=\mathcal{O}(x^{1/2+\varepsilon})$$. By the prime number theorem, $$\sum\limits_{n = 0}^\infty {\frac{{\mu (n)}}{n}} = 0.$$ Thus $$\sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} - \int_x^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} ,$$ where \begin{align*} \frac{{M(x)}}{x} - \int_x^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} & = \mathcal{O}(x^{ - 1/2 + \varepsilon } ) + \mathcal{O}(1)\int_x^{ + \infty } {\frac{{\rm d}t}{{t^{3/2 - \varepsilon } }}} \\ & = \mathcal{O}(x^{ - 1/2 + \varepsilon } ) + \mathcal{O}(x^{ - 1/2 + \varepsilon } ) = \mathcal{O}(x^{ - 1/2 + \varepsilon } ). \end{align*} Now assume $$G(x)=\mathcal{O}(x^{ - 1/2 + \varepsilon } )$$. Then, by Abel's summation formula, $$M(x):=\sum\limits_{n \le x} {\mu (n)} = \sum\limits_{n \le x} {\frac{{\mu (n)}}{n}n} = x\,G(x) + \int_1^x {G(t)\,{\rm d}t} .$$ Using our assumption, it follows that $$M(x)=\mathcal{O}(x^{1/2+\varepsilon})$$, which we know implies RH.

• At the third line, why do the integration limit changes? Commented Mar 16, 2023 at 12:32
• Because I added and subtracted $$\int_1^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} \quad \left( { = \sum\limits_{n = 0}^\infty {\frac{{\mu (n)}}{n}} = 0} \right).$$
– Gary
Commented Mar 16, 2023 at 12:33
• Thanks. For the fourth line, if $\epsilon>1/2$, isn't the integral diverges? Commented Mar 16, 2023 at 12:36
• You can assume that $0<\varepsilon<1/2$. Then the case $\varepsilon \ge 1/2$ follows trivially.
– Gary
Commented Mar 16, 2023 at 12:37
• No need of the PNT if you are assuming that $\sum_{n\ge 1}\mu(n)n^{-s}$ converges for $\Re(s) > 1/2$ as it implies that $\sum_{n\ge 1}\mu(n)n^{-1}=\lim_{s\to 1} \sum_{n\ge 1}\mu(n)n^{-s}=0$ :-) Commented Mar 16, 2023 at 13:17